A212427 -id:A212427 - cp's OEIS frontend

A212428 a(n) = 18*n + A000217(n-1).

0, 18, 37, 57, 78, 100, 123, 147, 172, 198, 225, 253, 282, 312, 343, 375, 408, 442, 477, 513, 550, 588, 627, 667, 708, 750, 793, 837, 882, 928, 975, 1023, 1072, 1122, 1173, 1225, 1278, 1332, 1387, 1443, 1500, 1558, 1617, 1677, 1738, 1800, 1863, 1927, 1992, 2058

Offset: 0

Jesse Han, May 16 2012

Generalization: T(n,i) = A000217(i-1+n) - A000217(i-1) = i*n + A000217(n-1) (corrected by Zak Seidov, Jun 21 2012); in this case is i=18.

For i = 11..16, Milan Janjic observed that if we define f(n,b,i) = Sum_{k=0..n-b} binomial(n,k)*Stirling1(n-k,b)*Product_{j=0..k-1} (-i - j), then T(n-1,i) = -f(n,n-1,i) for n >= 1.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A000096, A001008, A002805, A055998-A056000, A056115, A056119, A056121, A056126, A051942, A101859, A132754-A132758, A212427.

[n*(n+35)/2: n in [0..48]]; // Bruno Berselli, Jun 21 2012

Table[-18 (18 - 1)/2 + (18 + n) (17 + n)/2, {n, 0, 100}]
LinearRecurrence[{3,-3,1},{0,18,37},60] (* Harvey P. Dale, Jun 09 2024 *)

a(n)=n*(n+35)/2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = (17+n)*(18+n)/2 - 17*18/2 = 18*n + (n-1)*n/2 = n*(n+35)/2.
G.f.: x*(18-17*x)/(1-x)^3. - Bruno Berselli, Jun 21 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jul 10 2012
a(n) = 18*n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
From Amiram Eldar, Jan 11 2021: (Start)
Sum_{n>=1} 1/a(n) = 2*A001008(35)/(35*A002805(35)) = 54437269998109/229732925058000.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/35 - 102126365345729/2527062175638000. (End)
E.g.f.: exp(x)*x*(36 + x)/2. - Elmo R. Oliveira, Dec 12 2024

A303273 Array T(n,k) = binomial(n, 2) + k*n + 1 read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 4, 4, 1, 4, 6, 7, 7, 1, 5, 8, 10, 11, 11, 1, 6, 10, 13, 15, 16, 16, 1, 7, 12, 16, 19, 21, 22, 22, 1, 8, 14, 19, 23, 26, 28, 29, 29, 1, 9, 16, 22, 27, 31, 34, 36, 37, 37, 1, 10, 18, 25, 31, 36, 40, 43, 45, 46, 46, 1, 11, 20, 28, 35, 41

Offset: 0

Franck Maminirina Ramaharo, Apr 20 2018

Columns are linear recurrence sequences with signature (3,-3,1).
8*T(n,k) + A166147(k-1) are squares.
Columns k are binomial transforms of [1, k, 1, 0, 0, 0, ...].
Antidiagonals sums yield A116731.

			The array T(n,k) begins
1    1    1    1    1    1    1    1    1    1    1    1    1  ...  A000012
1    2    3    4    5    6    7    8    9   10   11   12   13  ...  A000027
2    4    6    8   10   12   14   16   18   20   22   24   26  ...  A005843
4    7   10   13   16   19   22   25   28   31   34   37   40  ...  A016777
7   11   15   19   23   27   31   35   39   43   47   51   55  ...  A004767
11  16   21   26   31   36   41   46   51   56   61   66   71  ...  A016861
16  22   28   34   40   46   52   58   64   70   76   82   88  ...  A016957
22  29   36   43   50   57   64   71   78   85   92   99  106  ...  A016993
29  37   45   53   61   69   77   85   93  101  109  117  125  ...  A004770
37  46   55   64   73   82   91  100  109  118  127  136  145  ...  A017173
46  56   66   76   86   96  106  116  126  136  146  156  166  ...  A017341
56  67   78   89  100  111  122  133  144  155  166  177  188  ...  A017401
67  79   91  103  115  127  139  151  163  175  187  199  211  ...  A017605
79  92  105  118  131  144  157  170  183  196  209  222  235  ...  A190991
...
The inverse binomial transforms of the columns are
1    1    1    1    1    1    1    1    1    1    1    1    1  ...
0    1    2    3    4    5    6    7    8    9   10   11   12  ...
1    1    1    1    1    1    1    1    1    1    1    1    1  ...
0    0    0    0    0    0    0    0    0    0    0    0    0  ...
0    0    0    0    0    0    0    0    0    0    0    0    0  ...
0    0    0    0    0    0    0    0    0    0    0    0    0  ...
...
T(k,n-k) = A087401(n,k) + 1 as triangle
1
1   1
1   2   2
1   3   4   4
1   4   6   7   7
1   5   8  10  11  11
1   6  10  13  15  16  16
1   7  12  16  19  21  22  22
1   8  14  19  23  26  28  29  29
1   9  16  22  27  31  34  36  37  37
1  10  18  25  31  36  40  43  45  46  46
...

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics: A Foundation for Computer Science, Addison-Wesley, 1994.

Cf. A085475, A086270, A086271, A086272, A086273, A130154, A159798, A162609, A162610, A300401.

T := (n, k) -> binomial(n, 2) + k*n + 1;
for n from 0 to 20 do seq(T(n, k), k = 0 .. 20) od;

Table[With[{n = m - k}, Binomial[n, 2] + k n + 1], {m, 0, 11}, {k, m, 0, -1}] // Flatten (* Michael De Vlieger, Apr 21 2018 *)

T(n, k) := binomial(n, 2)+ k*n + 1$
for n:0 thru 20 do
    print(makelist(T(n, k), k, 0, 20));

T(n,k) = binomial(n, 2) + k*n + 1;
tabl(nn) = for (n=0, nn, for (k=0, nn, print1(T(n, k), ", ")); print); \\ Michel Marcus, May 17 2018

G.f.: (3*x^2*y - 3*x*y + y - 2*x^2 + 2*x - 1)/((x - 1)^3*(y - 1)^2).
E.g.f.: (1/2)*(2*x*y + x^2 + 2)*exp(y + x).
T(n,k) = 3*T(n-1,k) - 3*T(n-2,k) + T(n-3,k), with T(0,k) = 1, T(1,k) = k + 1 and T(2,k) = 2*k + 2.
T(n,k) = T(n-1,k) + n + k - 1.
T(n,k) = T(n,k-1) + n, with T(n,0) = 1.
T(n,0) = A152947(n+1).
T(n,1) = A000124(n).
T(n,2) = A000217(n).
T(n,3) = A034856(n+1).
T(n,4) = A052905(n).
T(n,5) = A051936(n+4).
T(n,6) = A246172(n+1).
T(n,7) = A302537(n).
T(n,8) = A056121(n+1) + 1.
T(n,9) = A056126(n+1) + 1.
T(n,10) = A051942(n+10) + 1, n > 0.
T(n,11) = A101859(n) + 1.
T(n,12) = A132754(n+1) + 1.
T(n,13) = A132755(n+1) + 1.
T(n,14) = A132756(n+1) + 1.
T(n,15) = A132757(n+1) + 1.
T(n,16) = A132758(n+1) + 1.
T(n,17) = A212427(n+1) + 1.
T(n,18) = A212428(n+1) + 1.
T(n,n) = A143689(n) = A300192(n,2).
T(n,n+1) = A104249(n).
T(n,n+2) = T(n+1,n) = A005448(n+1).
T(n,n+3) = A000326(n+1).
T(n,n+4) = A095794(n+1).
T(n,n+5) = A133694(n+1).
T(n+2,n) = A005449(n+1).
T(n+3,n) = A115067(n+2).
T(n+4,n) = A133694(n+2).
T(2*n,n) = A054556(n+1).
T(2*n,n+1) = A054567(n+1).
T(2*n,n+2) = A033951(n).
T(2*n,n+3) = A001107(n+1).
T(2*n,n+4) = A186353(4*n+1) (conjectured).
T(2*n,n+5) = A184103(8*n+1) (conjectured).
T(2*n,n+6) = A250657(n-1) = A250656(3,n-1), n > 1.
T(n,2*n) = A140066(n+1).
T(n+1,2*n) = A005891(n).
T(n+2,2*n) = A249013(5*n+4) (conjectured).
T(n+3,2*n) = A186384(5*n+3) = A186386(5*n+3) (conjectured).
T(2*n,2*n) = A143689(2*n).
T(2*n+1,2*n+1) = A143689(2*n+1) (= A030503(3*n+3) (conjectured)).
T(2*n,2*n+1) = A104249(2*n) = A093918(2*n+2) = A131355(4*n+1) (= A030503(3*n+5) (conjectured)).
T(2*n+1,2*n) = A085473(n).
a(n+1,5*n+1)=A051865(n+1) + 1.
a(n,2*n+1) = A116668(n).
a(2*n+1,n) = A054569(n+1).
T(3*n,n) = A025742(3*n-1), n > 1 (conjectured).
T(n,3*n) = A140063(n+1).
T(n+1,3*n) = A069099(n+1).
T(n,4*n) = A276819(n).
T(4*n,n) = A154106(n-1), n > 0.
T(2^n,2) = A028401(n+2).
T(1,n)*T(n,1) = A006000(n).
T(n*(n+1),n) = A211905(n+1), n > 0 (conjectured).
T(n*(n+1)+1,n) = A294259(n+1).
T(n,n^2+1) = A081423(n).
T(n,A000217(n)) = A158842(n), n > 0.
T(n,A152947(n+1)) = A060354(n+1).
floor(T(n,n/2)) = A267682(n) (conjectured).
floor(T(n,n/3)) = A025742(n-1), n > 0 (conjectured).
floor(T(n,n/4)) = A263807(n-1), n > 0 (conjectured).
ceiling(T(n,2^n)/n) = A134522(n), n > 0 (conjectured).
ceiling(T(n,n/2+n)/n) = A051755(n+1) (conjectured).
floor(T(n,n)/n) = A133223(n), n > 0 (conjectured).
ceiling(T(n,n)/n) = A007494(n), n > 0.
ceiling(T(n,n^2)/n) = A171769(n), n > 0.
ceiling(T(2*n,n^2)/n) = A046092(n), n > 0.
ceiling(T(2*n,2^n)/n) = A131520(n+2), n > 0.

A358269 a(n) is the position m of the last prime term in the sequence {b(m)} defined by b(1) = n, if b(m) is prime then b(m+1) = b(m) - m, else b(m+1) = b(m) + m.

Original entry on oeis.org

3, 1004, 3, 1004, 3, 1004, 30, 349, 30, 5, 19, 5, 30, 1004, 30, 8, 11, 8, 30, 5, 86, 17, 67, 17, 15, 9, 19, 9, 15, 9, 19, 484, 19, 13, 30, 9, 19, 9, 19, 13, 374, 13, 19, 13, 11, 484, 86, 484, 19, 13, 67, 16, 19, 16, 19, 484, 374, 484, 19, 484, 374, 24, 19, 13

Offset: 0

Samuel Harkness, Nov 06 2022

nonn

A sequence {b(m)} is guaranteed to have no more primes when the m-th term "k" with value "s" is the sum of at least 3 consecutive positive integers where the sum is "s" and the last consecutive positive integer in the sum is k-1. Any number which is the sum of at least three consecutive positive integers is guaranteed to be composite. By the definition of the sequence, the next term k + 1 = s + k, and this term will be the sum of at least three consecutive positive integers with the last consecutive positive integer being k. This guarantees that this term is also guaranteed to be composite, and by induction, all future terms in {b(m)} will be composite.
In a sequence {b(m)}, if the m-th term k with value s satisfies c = (sqrt(-8*s + 4*k^2 - 4*k + 1) + 1)/2 for a positive integer c with s being nonprime and k > 3 then the value of all terms >= k will be composite.
It is unknown whether all initial conditions "n" guarantee a final prime. All terms up to n = 1000 have a final prime.
Treat negative numbers in the sequence {b(m)} as nonprime. The only n whose {b(m)} contain negative terms b(m) are 1, 3, 6, and 7.

			For n = 9: b(1) = 9. Nonprime, b(2) = 9 + 1 = 10. Nonprime, b(3) = 10 + 2 = 12. Nonprime, b(4) = 12 + 3 = 15. Nonprime, b(5) = 15 + 4 = 19. Prime, b(6) = 19 - 5 = 14. Note 14 = 2 + 3 + 4 + 5 and is nonprime, so b(7) = 2 + 3 + 4 + 5 + 6 and nonprime. All b(m) after this will be nonprime by the same pattern, thus the final prime for b(1) = 9 occurs at b(5), and a(9) = 5.

Samuel Harkness, Table of n, a(n) for n = 0..1111

Cf. A074171, A358166

Examples of sequences of the sum of consecutive positive integers, where the sum of at least three is guaranteed to be composite: A055999, A212427.

T = {}; For[f = 0, f <= 63, f++, a = 0; t = f; q = 0; While[a == 0, q++; If[t < 0, t += q, If[PrimeQ[t], t -= q; If[t >= 0, If[q != 2 && q != 1 && ! PrimeQ[t], s = t; k = q + 1; z = (Sqrt[-8 s + 4 k^2 - 4 k + 1] + 1)/2; If[Element[z, Reals] && z > 0 && Mod[z, 1] == 0, AppendTo[T, q]; Break[]]]], t += q]]]]; Print[T]

cp's OEIS Frontend

A212428 a(n) = 18*n + A000217(n-1).

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A303273 Array T(n,k) = binomial(n, 2) + k*n + 1 read by antidiagonals.

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A358269 a(n) is the position m of the last prime term in the sequence {b(m)} defined by b(1) = n, if b(m) is prime then b(m+1) = b(m) - m, else b(m+1) = b(m) + m.

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